The force F points in the direction of the unit vector e = 2/3 i -2/3 j +1/3 k .

Question

The force F points in the direction of the unit vector e = 2/3 i -2/3 j +1/3 k . the support at O with coordinates (.25 i +.2 j +.15 k ) will safely support a moment of 560 N-m magnitude. a) based on this criterion what is the largest safe magnitude of F? b) if the force F may be exerted in any direction, what is its largest safe magnitude? thank you ! The force F points in the direction of the unit vector e = 2/3 i -2/3 j +1/3 k . the support at O with coordinates (.25 i +.2 j +.15 k ) will safely support a moment of 560 N-m magnitude. a) based on this criterion what is the largest safe magnitude of F? b) if the force F may be exerted in any direction, what is its largest safe magnitude? thank you !

Explanation / Answer

(a)The unit vector is e = (2/3)i - (2/3)j + (1/3)k
The coordinates of the support at O is (.25i + .2j + .15k)
The position vector Oe is [(2/3)i - (2/3)j + (1/3)k - (.25i + .2j + .15k)] = (0.41i - 0.86j + 0.18k)
The length of the vector Oe is
|Oe| = [(0.41)^2 + (-0.86)^2 + (0.18)^2]^1/2 = [0.1681 + 0.7396 + 0.0324]^1/2 = (0.9401)^1/2 = 0.969 m
The magnitude of moment(torque) is
t = 560 N-m
We know from the relation t = F X |Oe| = F * |Oe| * sin? where ? is the angle between the force and perpendicular distance vectors The largest safe magnitude of F(? maximum = 90o) is t = F * |Oe| or F = (t/|Oe|) (b)When the force F is exerted in any direction,its largest safe magnitude is F = (t/|Oe| * sin?)

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